Need help with antiderivatives and integrals

Find f(x) given f"(x) = sin(x) + cos(x)
f(0) = 3
f(pi/2) = 4 + pi/8
Anyone know how to solve it? I'm pretty much stuck. I got that f'(x) = -cos(x) + sin(x) + C and
f(x) = -sin(x) - cos(x) + C + D but I don't know how to solve for C and D from the given information.

From the beginning:

if f ''(x)= sin(x) + cos(x)

then f '(x)= cos(x) - sin(x) + C

Now f (x)= -sin(x) - cos(x) + Cx + D

*Don't forget the antiderivative of a a constant C = Cx

Now go ahead and use the initial conditions:

f(0)=3= 0 -1 +D
D= 4

I am assuming you wrote f '(pi/2) = 4 + pi/8 , but I didn't see the prime.

f '(pi/2) = 4 + pi/8 = cos(pi/2) - sin(pi/2) + C
4 + pi/8 = 0 -1 + C
C= 5 + pi/8

Our final solution is thus:
f (x) = -sin(x) - cos(x) + (5 + pi/8)x + 4

Hope this helps :)